Question

1. a 2.00 × 10² -g sample of water at 80.0°c is mixed with 2.00 × 10² g of water at 10.0°c in a calorimeter. what is the final temperature of the mixture?
2. a 1.50 × 10² -g piece of glass at a temperature of 70.0°c is placed in a container with 1.00 × 10² g of water initially at a temperature of 16.0°c. what is the equilibrium temperature of the water?
3. challenge a 4.00 × 10² -g sample of water at 15.0°c is mixed with 4.00 × 10² g of water at 85.0°c. after the system reaches thermal equilibrium, 4.00 × 10² g of methanol at 15°c is added. assume there is no thermal energy lost to the surroundings. what is the final temperature of the mixture?

Explanation:

Problem 6

Step1: Define heat balance equation

Heat lost by hot water = Heat gained by cold water. Let final temp be $T_f$.
$\displaystyle m_1c(T_1-T_f) = m_2c(T_f-T_2)$

Step2: Cancel $c$ and substitute values

$c$ (specific heat of water) cancels out. $m_1=200\ \text{g}, T_1=80.0^\circ\text{C}; m_2=200\ \text{g}, T_2=10.0^\circ\text{C}$
$200(80.0-T_f) = 200(T_f-10.0)$

Step3: Simplify and solve for $T_f$

Divide both sides by 200: $80.0-T_f = T_f-10.0$
$2T_f = 90.0 \implies T_f = 45.0^\circ\text{C}$

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Problem 7

Step1: Define heat balance equation

Heat lost by glass = Heat gained by water. $c_{glass}=0.840\ \text{J/g}^\circ\text{C}$, $c_{water}=4.184\ \text{J/g}^\circ\text{C}$
$\displaystyle m_gc_g(T_g-T_f) = m_wc_w(T_f-T_w)$

Step2: Substitute given values

$m_g=150\ \text{g}, T_g=70.0^\circ\text{C}; m_w=100\ \text{g}, T_w=16.0^\circ\text{C}$
$150\times0.840\times(70.0-T_f) = 100\times4.184\times(T_f-16.0)$

Step3: Calculate constants and simplify

Left side: $126\times(70.0-T_f) = 8820 - 126T_f$
Right side: $418.4\times(T_f-16.0) = 418.4T_f - 6694.4$
$8820 + 6694.4 = 418.4T_f + 126T_f$

Step4: Solve for $T_f$

$15514.4 = 544.4T_f \implies T_f = \frac{15514.4}{544.4} \approx 28.5^\circ\text{C}$

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Problem 8

Step1: First mix water samples, find $T_{f1}$

Heat lost by hot water = Heat gained by cold water. $m_1=400\ \text{g}, T_1=85.0^\circ\text{C}; m_2=400\ \text{g}, T_2=15.0^\circ\text{C}$
$400(85.0-T_{f1}) = 400(T_{f1}-15.0)$

Step2: Solve for $T_{f1}$

$85.0-T_{f1}=T_{f1}-15.0 \implies 2T_{f1}=100.0 \implies T_{f1}=50.0^\circ\text{C}$
Total mass of water: $400+400=800\ \text{g}$

Step3: Mix with methanol, define heat balance

Heat lost by water = Heat gained by methanol. $c_{methanol}=2.53\ \text{J/g}^\circ\text{C}$, $m_m=400\ \text{g}, T_m=15^\circ\text{C}$
$\displaystyle m_wc_w(T_{f1}-T_f) = m_mc_m(T_f-T_m)$

Step4: Substitute values and solve

$800\times4.184\times(50.0-T_f) = 400\times2.53\times(T_f-15)$
$3347.2(50.0-T_f) = 1012(T_f-15)$
$167360 - 3347.2T_f = 1012T_f - 15180$
$167360+15180 = 1012T_f+3347.2T_f$
$182540 = 4359.2T_f \implies T_f = \frac{182540}{4359.2} \approx 41.9^\circ\text{C}$

Answer:

1. $45.0^\circ\text{C}$
2. $28.5^\circ\text{C}$
3. $41.9^\circ\text{C}$